Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{2z + 16}{z^2 - 12z + 27} \times \dfrac{6z - 18}{z + 8} $
Answer: First factor the quadratic. $t = \dfrac{2z + 16}{(z - 3)(z - 9)} \times \dfrac{6z - 18}{z + 8} $ Then factor out any other terms. $t = \dfrac{2(z + 8)}{(z - 3)(z - 9)} \times \dfrac{6(z - 3)}{z + 8} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ 2(z + 8) \times 6(z - 3) } { (z - 3)(z - 9) \times (z + 8) } $ $t = \dfrac{ 12(z + 8)(z - 3)}{ (z - 3)(z - 9)(z + 8)} $ Notice that $(z + 8)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ 12(z + 8)\cancel{(z - 3)}}{ \cancel{(z - 3)}(z - 9)(z + 8)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $t = \dfrac{ 12\cancel{(z + 8)}\cancel{(z - 3)}}{ \cancel{(z - 3)}(z - 9)\cancel{(z + 8)}} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $t = \dfrac{12}{z - 9} ; \space z \neq 3 ; \space z \neq -8 $